Hi, I’ve been stuck on this problem for a week and can’t seem to find any example that would help me to solve it. I would really appreciate it if someone would explain the steps to solving this problem. Thank you so much.
Hi, I’ve been stuck on this problem for a week and can’t seem to find any example that would help me to solve it. I would really appreciate it if someone would explain the steps to solving this problem. Thank you so much.
I developed a theorem (imcomplete) sometime ago, that if you have an integral linear combination of irrational distinct radicals equal to each other then there corresponding parts are equal.
For example, if
$\displaystyle 1+\sqrt{2}+\sqrt{3}=x+y\sqrt{2}+z\sqrt{3}$
where $\displaystyle x,y,z$ are integers then,
$\displaystyle x=1,y=1,z=1$.
Now if you have,
$\displaystyle 1+\sqrt{2}+\sqrt{3}=x+y\sqrt{5}$
where $\displaystyle x,y$ are integers is just impossible,
If you have that,
$\displaystyle \sqrt{16-6\sqrt{3}}=x+y\sqrt{3}$
where $\displaystyle x,y$ are integers then,
$\displaystyle 16-6\sqrt{3}=(x+y\sqrt{3})^2$
Thus,
$\displaystyle 16-6\sqrt{3}=(x^2+3y^2)+2xy\sqrt{3}$
Then,
$\displaystyle \left\{ \begin{array}{c}x^2+3y^2=16\\ 2xy=6$ is just impossible in integers.
This is my reasoning that why I would not expect a simpler solution with the integers.
No I did not make a mistake in what I said.Originally Posted by CaptainBlack
Note I made a mistake in doing
$\displaystyle \sqrt{16-6\sqrt{3}}$ the wrong problem.
If you used my method then you will see that it works
-------------------------
I base it on the fact that,
$\displaystyle a_0+a_1\sqrt{b_1}+...+a_n\sqrt{b_n}$
is always irrational whenver all are integers and the b_i are all non-squares and a_i are not zero.
Then given
$\displaystyle a_0+a_1\sqrt{b_1}+...+a_n\sqrt{b_n}$=$\displaystyle c_0+c_1\sqrt{b_1}+...+c_n\sqrt{b_n}$
You can write,
$\displaystyle d_0+d_1\sqrt{b_1}+...+d_n\sqrt{b_n}=0$ where $\displaystyle d_i=a_i-c_i$
Which is rational an impossibility unless a_i=c_i giving zeros.
You can use my method which I shown, (I only did the wrong example which is why it did not work).Originally Posted by Ranger SVO
You attempt to find integers $\displaystyle x,y$ such as,
$\displaystyle \sqrt{12-6\sqrt{3}}=x+y\sqrt{3}$
Thus,
$\displaystyle 12-6\sqrt{3}=(x+y\sqrt{3})^2$
Thus,
$\displaystyle 12-6\sqrt{3}=(x^2+3y^2)+2xy\sqrt{3}$
Then, by my previous post,
$\displaystyle \left\{ \begin{array}{c}12=x^2+3y^2\\ -6=2xy$
Look at the second equation,
$\displaystyle -6=2xy$
This, is equivalent to
$\displaystyle xy=-3$
Since $\displaystyle x,y$ are integers the only possibilities are,
$\displaystyle (x,y)=(-1,3),(3,-1),(-3,1),(1,-3)$
Now we check which one of these satisfies the second equation,
we see that,
$\displaystyle \begin{array}{c} (3)^2+3(-1)^2=12\\ (-3)^2+3(1)^2=12$
Thus,
$\displaystyle 3-\sqrt{3}$
$\displaystyle \sqrt{3}-3$
Both of these are true, indeed,
$\displaystyle (\sqrt{3}-3)^2=(3-\sqrt{3})^2=12-6\sqrt{3}$
But we need to chose,
$\displaystyle 3-\sqrt{3}$
Because square roots are non-negative and,
$\displaystyle \sqrt{3}-3<0$